Asked by madhubala
The potential of hydrogen electrode is -118MV.thr H^+ concentration of the solution is
a)0.01M
b)2M
C)10^4M
d)1M
a)0.01M
b)2M
C)10^4M
d)1M
Answers
Answered by
DrBob222
I assume you mean -118 millivolts and that pH2 = 1 atmosphere..
-0.118 v = Eo - (0.06/n)log (H2/(H^+)^2 for the equation
2H^+ 2e ==> H2
Post your work if you get stuck.
-0.118 v = Eo - (0.06/n)log (H2/(H^+)^2 for the equation
2H^+ 2e ==> H2
Post your work if you get stuck.
Answered by
Shraddha patel
Taking E =118millivolt...
2H+. +2e ...H2
E= E*-0.059/2log H2/[H+]2
-118=0-0.059/2log 1/[H+]2
Solving it by using antilog we will get 0.01
2H+. +2e ...H2
E= E*-0.059/2log H2/[H+]2
-118=0-0.059/2log 1/[H+]2
Solving it by using antilog we will get 0.01
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.